The dimensions of loss can be expressed in units of length*length/time*time (L2/
ID: 2991600 • Letter: T
Question
The dimensions of loss can be expressed in units of length*length/time*time (L2/t2). As such, the loss is a measure of:
how far the fluid can flow without friction.
the energy gain per unit mass due to friction.
the energy loss per unit mass due to friction.
how high the fluid can flow without friction.
The steady state energy equation with head loss (hL) and shaft work (hs) is written as
If the left hand side of the energy equation were larger than the right hand side:
the flow would come to a new equilibrium.
the flow would not be possible.
the flow would stop.
the flow would speed up.
the flow would slow down.
Water flows steadily through a smooth pipe with a Reynolds number Re1. If the velocity is doubled the relation between the Reynolds number Re2 after the increase to Re1 is
The change in Reynolds number depends on whether the flow is laminar or turbulent.
Water flows steadily through a smooth pipe. If the velocity is increased
the head loss stays the same.
the change in head loss depends on whether the flow is laminar or turbulent.
the head loss increases.
the head loss decreases.
the flow would come to a new equilibrium.
Explanation / Answer
Following statements are true and their reasoning provided: 1. The dimensions of loss can be expressed in units of length*length/time*time (L2/t2). As such, the loss is a measure of: the energy loss per unit mass due to friction. 2. The steady state energy equation with head loss (hL) and shaft work (hs) is written as If the left hand side of the energy equation were larger than the right hand side: the flow would speed up. 3. Water flows steadily through a smooth pipe with a Reynolds number Re1. If the velocity is doubled the relation between the Reynolds number Re2 after the increase to Re1 is: Re2 = 2*Re1 4. Water flows steadily through a smooth pipe. If the velocity is increased: the head loss increases. 5. A valve has a loss coefficient Kc. The valve is improved so that the loss coefficient is reduced to one-half the value. For the same flow rate, the relation between the pressure drop across the valve after the improvement p2 relative to that before ( p1) is: p2 = 0.5*p1